A car with a mass of 2.3 t moves evenly on a horizontal road. Determine the specific heat of combustion

A car with a mass of 2.3 t moves evenly on a horizontal road. Determine the specific heat of combustion of gasoline if, to cover 142 km of path, the car engine with an average resistance force equal to 0.03 of the weight of the car consumed 15 liters of fuel. The engine efficiency is 20%

To find the calculated value beats. the calorific value of gasoline, we use the formula: η = Ap / Q = F * S / (qb * mb) = μ * P * S / (qb * V * ρb) = μ * ma * g * S / (qb * V * ρb ), whence we express: qb = μ * ma * g * S / (η * V * ρb).

Variables and constants: μ – coeff. resistance (μ = 0.03); ma is the mass of the presented car (ma = 2.3 t = 2.3 * 10 ^ 3 kg); g – acceleration due to gravity (g ≈ 10 m / s2); S – distance traveled (S = 142 km = 142 * 10 ^ 3 m); η – engine efficiency (η = 20% = 0.2); V is the volume of consumed gasoline (V = 15 l = 15 * 10 ^ -3 m3); ρb is the density of gasoline (ρb = 710 kg / m3).

Calculation: qb = μ * ma * g * S / (η * V * ρb) = 0.03 * 2.3 * 10 ^ 3 * 10 * 142 * 10 ^ 3 / (0.2 * 15 * 10 ^ – 3 * 710) = 46 * 10 ^ 6 J / kg.

Answer: Ud. the heat of combustion of gasoline, according to the calculation, is equal to 46 MJ / kg.